Page 64 - PRE-U STPM MATHEMATICS (T) TERM 1

Mathematics Term 1 STPM Answers
π π 5. ±(1 + 2i), ±(1 – 2i)
3. (a) 2 3 cos + i sin 4 , 4 + 4i
4 4 π π π π 1
π
π
4
4
4
4
(b) 2 3 3 cos + i sin 4 ; –144( 3 – 3i) 6. (a) 2 1 cos + i sin 2 , 2 3 cos 1 – 2 + i sin 1 – 24 ; , π
2
3
3
6
8
1
π
π
(c) cos 1 – 2 + i sin 1 – 2 , – ( 3 + i) (b) 5 + i ; 2, 0.927 rad
5
6 6 2
5
(d) 13 cos (2.4) + i sin (2.4)], 13 [cos (5.88) + i sin (5.88)] 7. ±1.23 rad.
π
π
3
5
1
7
(e) 4 3 cos 1 – 2 + i sin 1 – 24 , 512(1 + 3i) 8. sin π, sin π, sin π, – 2 + 2
3 3 8 8 8 2
(f) 23 [cos (0.485) + i sin (0.485)], 1 1

2
2
— 5 9. (a) a = (3 + 21), b = (–3 + 21)
2
23 [cos (2.425) + i sin (2.425)] (b) (i) –1, – 3 (ii) 5 , π
4. (a) cos 23 q + i sin 23 q (b) cos 8q + i sin 8q 4 12
12 12 1 3 1 3
2
(c) cos 5q + i sin 5q (d) cos 29 q + i sin 29 q 10. 4 cos q – 2; ±i, + 2 i, – 2 i
2
2
12
12
(e) cos 11 q + i sin 11 q π π π π

12 12 11. 2 3 cos + i sin 4 , 2 2 3 cos 1 – 2 + i sin 1 – 24 ;
4
6
4
6
π
π
5
5
5. 1 3 cos + i sin 4 , i 1 2k + 2 π + i sin 1 2k + 2 π4 , k = 0, 1, 2
2 4 4 32 — 1 3 cos 1 3 36 3 36
— 1 π π 2 6
4
6. (a) ±2 3 cos + i sin 4
8 8 13. a = 5, b = –10, c = 1;
(b) ± 5 [cos (–0.464) + i sin (–0.464)] 2 cos , 2 cos π, 2 cos π, 2 cos π
3
4
2
π
(c) ± 13 [cos (–0.588) + i sin (–0.588)] 5 5 π 5 5
π
π
(d) ± 2 3 cos 1 – 2 + i sin 1 – 24 14. (a) |z| = 4 ; arg z = – 6
12
12
— 3 π π (c) 1 (–1 + 3i)
4
(e) ±2 3 cos 1 – 2 + i sin 1 – 24 32
8
8
2π
— 1 4 π π 15. 83 cos 2π + i sin 4 ;
(f) ±3 3 cos + i sin 4 3 3
8 8
— 1 2k 1 2k 1 1.5321 + 1.2856i, –1.8794 + 0.6840i, 0.3473 – 1.9696i
7. (a) 2 3 cos 1 + 2 π + i sin 1 + 2 π4 , k = 0, 1, 2
6
3 12 3 12 16. (a) w = 1 + i ; real part = 1, imaginary part = 1
π
π
1
—
(b) 5 3 cos 1 2k π – 0.3092 + i sin 1 2k – 0.30924 , k = 0, 1, 2 (b) w = 2 3 cos + i sin 4
3
4
4
3
3
— 1 2k 2k (c) 1.0842 + 0.2905i, –0.7937 + 0.7937i, –0.2905 – 1.0842i
(c) 13 3 cos 1 π – 0.3922 + i sin 1 π – 0.39224 , k = 0, 1, 2
3
3 3 π π 5π 5π
— 1 2k 1 2k 1 17. z = 4 2 3 cos1 – 2 + i sin 1 – 24 , w = 2 3 1 cos + i sin 4 ;
(d) 2 3 cos 1 3 – 2 π + i sin 1 3 – 2 π4 , k = 0, 1, 2 8 4 4 6 6
3
18
18
π
π
— 1 2k 2k w 6 = 81 3 cos + i sin 4
1
1
(e) 2 3 cos 1 – 2 π + i sin 1 – 2 π4 , k = 0, 1, 2 z 128 6 6
3
3 12 3 12
— 1 2k 1 2k 1 18. (a) 3 – i, 1 + i 3, – 3 + i, –1 – i 3
(f) 3 3 cos 1 + 2 π + i sin 1 + 2 π4 , k = 0, 1, 2
6
3 12 3 12 19. (a) real part = 32, imaginary part = –32 3
nπ
π
n
8. |z| = 2, arg z = , z = 2 3 cos nπ + i sin 4 π π
n
3 3 3 (b) z = 83 cos 1 – 2 + i sin 1 – 24 = 4 3 – 4i
1
1 2 2 1 6 6
2
3
9. z = 3 cos π + i sin π4 , z = [cos π + i sin π] 5π 5π
4 3 3 8 z = 83 cos + i sin 4 = –4 3 + 4i
6
6
2
π
π
3
3
10. 23 cos + i sin 4 , 2 2 3 cos π + i sin π4
6 6 4 4
7
7

z = 1 1 3 cos 1 2k – 2 π + i sin 1 2k – 2 π4 , k = 0, 1, 2 Chapter 5 Analytic Geometry
— 3 36 3 36
2 6 Exercise 5.1
π
2
2
11. q = , π 1. 3x + 3y – 16y + 16 = 0
3 4 2. y = 3a(2x – 3a)
2
2
2
STPM PRACTICE 4 3. x + y + 4y = 0
4. 16x + 10y – 15 = 0
7
4
1. (a) p = , q = – 5. x + y – 5x = 0
2
2
5 5
2
(b) p = 2 – i, q = 2 + i 6. (a) (y – 3) = x + 1
2
(b) 27y = 4x 3
3. (1, 2), (–1, –2) (c) (x – 4)y = 9
1
4. |z | = 2 , arg z = – π (d) y = (x – 1)(x – 2)
2
1 1 4 (e) x = 12(y – 1)
3
2
|z | = 2 2 , arg z = – π (f) (y – 1) = (x + 1) 3
4
2
2
(g) y(x – 1) = 8
2
|z | = 1, arg z = – π (h) x (y + 1) = 64
2
3 3 3
283
Answers STPM Math T S1.indd 283 3/28/18 4:25 PM